The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ is

The radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$. Then

If the circle $x^{2}+y^{2}+2gx+2fy+c=0$ cuts the three circles $x^{2}+y^{2}-5=0$, $x^{2}+y^{2}-8x-6y+10=0$, and $x^{2}+y^{2}-4x+2y-2=0$ at the extremities of their diameters, then

Find the equation of the circle passing through the point $(1, 1)$ and the intersection points of the circles $x^2 + y^2 = 6$ and $x^2 + y^2 - 6x + 8 = 0$.

$A$ circle $C$ touches the line $x=2y$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2y-5=0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1}$. Then the diameter of $C$ is:

The line $x-2=0$ cuts the circle $x^2+y^2-8x-2y+8=0$ at $A$ and $B$. The equation of the circle passing through the points $A$ and $B$ and having the least radius is